[SOLVED] Seeking a better, more accurate and more elegant proof

Let be the set of all nilpotent elements of a commutative ring , and show that for each prime ideal of .

Proof: Let and . Then there is a positive integer with . Assume towards a contradiction that . Since and , then by definition of prime ideals we have . However, the same can be said for any integer , that is, if then and therefore . By induction, since , then , which contradicts our assumption that . So whenever , that is, .

The reasoning is sound, but I'm not quite sure how to best articulate it. I don't think "by induction" is completely appropriate, here. Also, I like to avoid by-contradiction proofs if possible. Does anyone have an idea for how I can more elegantly and accurately explain what's going on in this exercise?

Proof: Let and . Then there is a positive integer with . Assume towards a contradiction that . Since and , then by definition of prime ideals we have . However, the same can be said for any integer , that is, if then and therefore . By induction, since , then , which contradicts our assumption that . So whenever , that is, .

The reasoning is sound, but I'm not quite sure how to best articulate it. I don't think "by induction" is completely appropriate, here. Also, I like to avoid by-contradiction proofs if possible. Does anyone have an idea for how I can more elegantly and accurately explain what's going on in this exercise?

Thanks!

Why did you have to use contradiction at all? Let be any nilpotent, then (as you showed this is fine)...and voila! No contradiction and short.

If you don't like that "finite induction" argument, you can derive the result about nilpotent elements as a corollary of a more general result:

If a product lies in a prime ideal P, then at least one of the elements belongs to P.

That result is easily proved, for all , by a (standard) induction argument, the base case n=2 being just the definition of primality. The result about nilpotents follows by taking all the to be equal to .